Engineering Mechanics: Statics

Chapter 2: Force Systems Part A: Two Dimensional Force Systems

Force


Force = an action of one body on another Vector quantity




External and Internal forces




Mechanics of Rigid bodies: Principle of Transmissibility




• Specify magnitude, direction, line of action • No need to specify point of application


Concurrent forces • Lines of action intersect at a point

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2D Force Systems


Rectangular components are convenient for finding v the sum or resultant R of two (or more) forces which are concurrent v v v R = F1 + F2 = (F1xˆi + F1y ˆj ) + (F2 xˆi + F2 y ˆj ) = (F1x + F2 x )ˆi + (F1y + F2 y )ˆj

Actual problems do not come with reference axes. Choose the most convenient one!

Moment








In addition to tendency to move a body in the direction of its application, a force tends to rotate a body about an axis. The axis is any line which neither intersects nor is parallel to the line of action This rotational tendency is known as the moment M of the force  Proportional to force F and the perpendicular distance from the axis to the line of action of the force d  The magnitude of M is

M = Fd

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Moment





The moment is a vector M perpendicular to the plane of the body. Sense of M is determined by the right-hand rule  Direction of the thumb = arrowhead  Fingers curled in the direction of the rotational tendency




In a given plane (2D),we may speak of moment about a point which means moment with respect to an axis normal to the plane and passing through the point.




+, - signs are used for moment directions – must be consistent throughout the problem!

Moment





A vector approach for moment calculations is proper for 3D problems. Moment of F about point A maybe represented by the cross-product

M=rxF where r = a position vector from point A to any point on the line of action of F

M = Fr sin α = Fd

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Example 2/5 (p. 40) Calculate the magnitude of the moment about the base point O of the 600-N force by using both scalar and vector approaches.

Problem 2/50 (a) Calculate the moment of the 90-N force about point O for the condition θ = 15º. (b) Determine the value of θ for which the moment about O is (b.1) zero (b.2) a maximum

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Couple


Moment produced by two equal, opposite, and noncollinear forces = couple

M = F(a+d) – Fa = Fd 




Moment of a couple has the same value for all moment center

Vector approach

M = rA x F + rB x (-F) = (rA - rB) x F = r x F 

Couple M is a free vector

Couple


Equivalent couples  Change of values F and d  Force in different directions but parallel plane  Product Fd remains the same

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Force-Couple Systems



Replacement of a force by a force and a couple Force F is replaced by a parallel force F and a counterclockwise couple Fd

Example Replace the force by an equivalent system at point O Also, reverse the problem by the replacement of a force and a couple by a single force

Problem 2/76 (modified) The device shown is a part of an automobile seat-back-release mechanism. The part is subjected to the 4-N force exerted at A and a 300-N-mm restoring moment exerted by a hidden torsional spring. Find an equivalent force-couple system at point O of the 4-N force

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Resultants


The simplest force combination which can replace the original forces without changing the external effect on the rigid body




Resultant = a force-couple system

v v v v v R = F1 + F2 + F3 + K = ΣF Rx = ΣFx , Ry = ΣFy , R = (ΣFx )2 + (ΣFy )2

θ = tan-1

Ry Rx

Resultants


Choose a reference point (point O) and move all forces to that point




Add all forces at O to form the resultant force R and add all moment to form the resultant couple MO




Find the line of action of R by requiring R to have a moment of MO

v v R = ΣF MO = ΣM = Σ(Fd) Rd = MO

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Problem 2/87 Replace the three forces acting on the bent pipe by a single equivalent force R. Specify the distance x from point O to the point on the x-axis through which the line of action of R passes.

Problem 2/76 The device shown is a part of an automobile seat-back-release mechanism. The part is subjected to the 4-N force exerted at A and a 300-N-mm restoring moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the single equivalent force.

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Force Systems

Part B: Three Dimensional Force Systems

Three-Dimensional Force System


Rectangular components in 3D

• Express in terms of unit vectors

ˆi, ˆj, kˆ

v F = Fxˆi + Fy ˆj + Fz kˆ Fx = F cos θ x ,

Fy = F cos θy , Fz = F cos θ z

F = Fx 2 + Fy 2 + Fz 2

• cosθx, cosθy , cosθz are the direction cosines • cosθx = l, cosθy = m, cosθ z= n v F = F(liˆ + mjˆ + nkˆ)

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Three-Dimensional Force System


Rectangular components in 3D

• If the coordinates of points A and B on the line of action are known, v v v (x − x1)ˆi + (y 2 − y1)ˆj + (z2 − z1)kˆ AB F = FnF = F =F 2 AB (x2 − x1)2 + (y 2 − y1)2 + ( z2 − z1)2

• If two angles θ and φ which orient the line of action of the force are known,

Fxy = F cos φ ,

Fz = F sinφ

Fx = F cos φ cos θ ,

Fy = F cos φ sinθ

Problem 2/98


The cable exerts a tension of 2 kN on the fixed bracket at A. Write the vector expression for the tension T.

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Three-Dimensional Force System


Dot product

v v P ⋅ Q = PQ cos α  

Orthogonal projection of Fcosα of F in the direction of Q Orthogonal projection of Qcosα of Q in the direction of F

v v


We can express Fx = Fcosθx of the force F as Fx = F ⋅ i




If the projection of F in the n-direction is F ⋅ n

v v

Example


Find the projection of T along the line OA

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Moment and Couple


Moment of force F about the axis through point O is

MO = r x F   




r runs from O to any point on the line of action of F Point O and force F establish a plane A The vector Mo is normal to the plane in the direction established by the right-hand rule

Evaluating the cross product

ˆi MO = rx

ˆj ry

kˆ rz

Fx

Fy

Fz

Moment and Couple


Moment about an arbitrary axis

v v v v v Mλ = (r × F ⋅ n)n known as triple scalar product (see appendix C/7)


The triple scalar product may be represented by the determinant

rx v Mλ = Mλ = Fx l

ry

rz

Fy

Fz

m n

where l, m, n are the direction cosines of the unit vector n

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Sample Problem 2/10 A tension T of magniture 10 kN is applied to the cable attached to the top A of the rigid mast and secured to the ground at B. Determine the moment Mz of T about the z-axis passing through the base O.

Resultants


A force system can be reduced to a resultant force and a resultant couple

v v v v v R = F1 + F2 + F3 L = ∑ F v v v v v v M = M1 + M2 + M3 + L = ∑(r × F)

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Problem 2/154


The motor mounted on the bracket is acted on by its 160-N weight, and its shaft resists the 120-N thrust and 25-N.m couple applied to it. Determine the resultant of the force system shown in terms of a force R at A and a couple M.

Wrench Resultants


Any general force systems can be represented by a wrench

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Problem 2/143





Replace the two forces and single couple by an equivalent force-couple system at point A Determine the wrench resultant and the coordinate in the xy plane through which the resultant force of the wrench acts

Resultants


Special cases • Concurrent forces – no moments about point of concurrency • Coplanar forces – 2D • Parallel forces (not in the same plane) – magnitude of resultant = algebraic sum of the forces • Wrench resultant – resultant couple M is parallel to the resultant force R • Example of positive wrench = screw driver

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Problem 2/151





Replace the resultant of the force system acting on the pipe assembly by a single force R at A and a couple M Determine the wrench resultant and the coordinate in the xy plane through which the resultant force of the wrench acts

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